Question

ABCD is a rectangle of dimensions 80 cm \(\times\) 60 cm. Another rectangle PQRS is drawn inside ABCD leaving space of equal width x cm along the edges of ABCD. If area PQRS is half of the area ABCD, then find the value of x.

Hint:

Always double-check your final values against physical constraints.
The margin \(x\) cannot exceed half of the shortest side of the rectangle. Since the shortest side is 60 cm, \(x\) must be strictly less than 30 cm. This helps you instantly rule out the root \(x = 60\) without wasting time.

Answer 1

Correct Answer: 10

Step 1: Understand the Setup.
Rectangle $ABCD$ has dimensions $80$ cm $\times$ $60$ cm. Rectangle $PQRS$ is drawn inside it, leaving a uniform border of width $x$ cm on all four sides. The area of $PQRS$ is half the area of $ABCD$.
Step 2: Express the Dimensions of PQRS.
Since $x$ cm is removed from both sides (left and right), the length of $PQRS$ is $(80 - 2x)$ cm. Similarly, the width is $(60 - 2x)$ cm.
Step 3: Set Up the Area Equation.
Area of $ABCD = 80 \times 60 = 4800$ cm$^2$. Area of $PQRS = \dfrac{1}{2} \times 4800 = 2400$ cm$^2$. So: \[ (80 - 2x)(60 - 2x) = 2400 \]
Step 4: Expand and Simplify.
\[ 4800 - 160x - 120x + 4x^2 = 2400 \] \[ 4x^2 - 280x + 4800 - 2400 = 0 \] \[ 4x^2 - 280x + 2400 = 0 \] Dividing throughout by 4: \[ x^2 - 70x + 600 = 0 \]
Step 5: Solve the Quadratic Equation.
Using the quadratic formula: \[ x = \frac{70 \pm \sqrt{70^2 - 4 \times 600}}{2} = \frac{70 \pm \sqrt{4900 - 2400}}{2} = \frac{70 \pm \sqrt{2500}}{2} = \frac{70 \pm 50}{2} \] This gives $x = \dfrac{70+50}{2} = 60$ or $x = \dfrac{70-50}{2} = 10$.
Step 6: Select the Valid Solution.
If $x = 60$, then the width of $PQRS$ becomes $60 - 120 < 0$, which is impossible. So $x = 10$ cm. \[ \boxed{x = 10 \text{ cm}} \]